Return-Path: murayama@hitoshi.lbl.gov
Return-Path:
Received: from hitoshi.lbl.gov (localhost [127.0.0.1])
by hitoshi.lbl.gov (8.8.5/8.8.5) with ESMTP id LAA24304;
Thu, 30 Oct 1997 11:36:38 -0800
Message-Id: <199710301936.LAA24304@hitoshi.lbl.gov>
X-Mailer: exmh version 1.6.9 05/05/96
To: "Gary L. Leung"
cc: murayama@physics.berkeley.edu, murayama@hitoshi.lbl.gov
Subject: Re: Midterm problem
Reply-to: murayama@lbl.gov
In-reply-to: Your message of "Thu, 30 Oct 1997 02:06:48 PST."
Mime-Version: 1.0
Content-Type: text/plain; charset=us-ascii
Date: Thu, 30 Oct 1997 11:36:38 -0800
From: Hitoshi Murayama
In message , "Gary
L. Leung" writes:
>--MimeMultipartBoundary
>Content-Type: TEXT/PLAIN; charset=US-ASCII
>
>Prof Murayama,
>
>On Wed, 29 Oct 1997, Hitoshi Murayama wrote:
>
>> No. For a two-pion state, one needs to specify both the total isospin I and
>
>> the relative angular momentum L. It turns out that not all combinations of
>I
>> and L are allowed; the Bose statistics of the pion requires that the
>> wavefunction does not change when you exchange two pions, and it leaves only
>
>> particular combinations of I and L allowed. You are asked to determine what
>
>> combinations are allowed, or in other words, what values of L are allowed fo
>r
>> I=0 and I=1.
>
>I am still confused.
>For example if we want to create a two-nucleon state with proton (I=1/2)
>and neutron (I=1/2), the total isospin will either be I = 1 = 1/2 + 1/2 or
>I = 0 = 1/2 - 1/2. For I = 1, we can have pp, pn+np, nn. and for I = 0,
>we will have pn-np.
>this problem asks me to create a two-pion state with the three pions,
>namely pi+, pi0, and pi-, each with ( I = 1). So the total isospin
>should either be I = 2 = 1+1 or I = 0 = 1 - 1 if we are creating a
>two-pion state. I mean the state will be like |2,2> |2,1> |2,0> |2,-1>
>|2,-2>. So am I missing something here?
What you are missing is the rule when you add two angular momenta. With j_1
and j_2, you have the following possibilities for the total angular momentum:
J = |j_1 - j_2|, |j_1 - j_2|+1, ... , j_1 + j_2 - 1, j_1 + j_2 .
For instance with j_1=2 and j_1=1, the possibilities are J=1, 2, 3. Your
argument says only J=1 and 3 would be possible, but this is not true. You can
obtain J=2 as well. Similarly, when you add I_1=1 and I_2=1, there is another
possibility for the total isospin beyond I=0 and 2 you mentioned.
>>
>> > for 3b, I looked up the booklet and find out that the quark contents
>> >for p0 and pi0 are exactly the same. So why are they two different
>> >particles?
>>
>> I assume "p0" you refer to is rho0, where rho is a Greek letter.
> They are different particles because they have different spin!
> rho0 has spin 1, pi0 has spin 0.
>>
>I looked up all the conservation laws which might forbit the process,
>rho0 -> pi0 + pi0 but couldn't find one. Could you give me some hints on
>what I should be looking for?
This again comes from construction of wave functions. If you deal with the
wave functions for both isospin and relative orbital angular momentum
correctly, you are supposed to find that there is no allowed two-pion wave
function consistent with all conservation laws and Bose statistics of the
pions.
--
Hitoshi Murayama
Assistant Professor of Physics
University of California
Berkeley, CA 94720